If the locus of the mid-point of the line segment from the point (3,2) to a point on the c…

Question

If the locus of the mid-point of the line segment from the point $(3,2)$ to a point on the circle $x^2+y^2=1$ is a circle of radius $r$, then $r$ is equal to

Accepted Answer

Let P(h, k) and point on the circle is (cos$$\theta$$, sin$$\theta$$)

$$ \therefore $$ $${{3 + \cos \theta } \over 2} = h$$ and $${{2 + \sin \theta } \over 2} = k$$

cos$$\theta$$ = 2h - 3 and sin$$\theta$$ = 2h - 2

Squaring and adding we get

${(2h - 3)^2} + {(2h - 2)^2} = 1$

$$ \Rightarrow 4{x^2} - 12x + 9 + 4{y^2} - 8y + 4 = 1$$

$$ \Rightarrow 4{x^2} + 4{y^2} - 12x - 8y + 12 = 0$$

$$ \Rightarrow {x^2} + {y^2} - 3x - 2y + 3 = 0$$

Radius = $$\sqrt {{9 \over 4} + 1 - 3}  = {1 \over 2}$$

Solutions · 1